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3 years ago

Can anyone help me with this last question?

Physics

1 answer:

Digiron [165]3 years ago

6 0

Answer:

v = 39.6 m/s (first answer option)

Explanation:

Use conservation of mechanical energy.

The starting is completely potential energy and the final is all kinetic energy:

$m*g*h=\frac{1}{2}*m*v^2\\g * h=\frac{1}{2} v^2\\9.8*80=\frac{1}{2} v^2\\v^2=1568\\v\approx 39.6\,\,m/s$

which agrees with the first answer shown.

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Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

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Substitute all the parameters into the formula

10 × 5 - 10 × 3

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Remember KE=M*V2/2…  A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy

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EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

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Explanation:

According to the Bernoulli’s equation:

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∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

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$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

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J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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